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National Math Olympiad in Slovenia

Geometry Difficulty 4.7 AIME Prove it Slovenia

The bisector of the inner angle $\angle ACB$ in an acute triangle $ABC$ intersects the side $AB$ at $D$ . The circumcircle of the triangle $ADC$ intersects the side $BC$ in two distinct points, $C$ and $E$ . The line through $B$ parallel to the line $AE$ intersects the line $CD$ at $F$ . Prove that $AFB$ is an isosceles triangle.

Solution

Write $\angle ACB = \gamma$ . Then $\angle ACF = \angle FCB = \frac{\gamma}{2}$ , since $CF$ is the bisector of the angle $\angle ACB$ . The points $A$ , $D$ , $E$ and $C$ are concyclic, so $\angle DAE = \angle DCE = \frac{\gamma}{2}$ . The line $BF$ is parallel to the line $AE$ , so $\angle ABF = \angle DAE = \frac{\gamma}{2}$ . We have $\angle ACF = \frac{\gamma}{2} = \angle ABF$ , so the points $A$ , $F$ , $B$ and $C$ are concyclic. From here we get $\angle FAB = \angle FCB = \frac{\gamma}{2} = \angle ABF$ and so $AFB$ is an isosceles triangle with the apex at $F$ .

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